# If 8.40 kJ of heat is needed to raise the temperature of a sample of metal from 15 °C to 20 °C, how many kilojoules of heat will be required to raise the temperature of the same sample of metal from 25 °C to 40 °C?

Dec 29, 2017

$\text{25 kJ}$

#### Explanation:

The trick here is to realize that because the sample of metal has the same mass in both cases, you can say that

${q}_{2} = \frac{\Delta {T}_{2}}{\Delta {T}_{1}} \cdot {q}_{1}$

Here

• ${q}_{1}$ is the amount of heat needed to raise the temperature of the sample by $\Delta {T}_{1} = {20}^{\circ} \text{C" - 15^@"C}$
• ${q}_{2}$ is the amount of heat needed to raise the temperature of the sample by $\Delta {T}_{2} = {40}^{\circ} \text{C" - 25^@"C}$

This equation can be found by using the fact that the heat absorbed by the metal can be calculated using the equation

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{q = m \cdot c \cdot \Delta T}}}$

Here

• $m$ is the mass of the sample
• $c$ is the specific heat of the metal

In your case, you can say that

${q}_{1} = m \cdot c \cdot \Delta {T}_{1}$

and

${q}_{2} = m \cdot c \cdot \Delta {T}_{2}$

Divide these two equations

${q}_{1} / {q}_{2} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{m \cdot c}}} \cdot \Delta {T}_{1}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{m \cdot c}}} \cdot \Delta {T}_{2}}$

to get

${q}_{2} = \frac{\Delta {T}_{2}}{\Delta {T}_{1}} \cdot {q}_{1}$

This equation tells you that in order to increase the temperature of the metal by a factor $\frac{\Delta {T}_{2}}{\Delta {T}_{1}}$ when the mass of the metal is constant, the amount of heat supplied must also increase by a factor of $\frac{\Delta {T}_{2}}{\Delta {T}_{1}}$.

So, plug in your values to find

q_2 = ((40 - 25) color(red)(cancel(color(black)(""^@"C"))))/((20-15)color(red)(cancel(color(black)(""^@"C")))) * "8.40 kJ"

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{q}_{2} = \text{25 kJ}}}}$

The answer is rounded to two sig figs.